Exploring Electron Triplet States for Physics Questions

[allisonpw]

How did you find PF?: internet question about electron triplet states

I'm a retired accelerator physicist entertaining myself by trying to understand physics questions like entanglement better. I calculate the expectation value of the product of two electron spins in the singlet state with measurement #1 at arbitrary angle phi wrt z-axis, #2 measurement angle phi+ theta, getting the -cos(theta) answer independent of phi that I expected, as per Bell's theorem experiments. But with Susskind's ("Quantum Mechanics Theoretical Minimum") maximally entangled triplet states, I get either cos(theta) or cos (theta+2*phi). I'm doubtful that my results are correct. What are the answers and where are they worked out?

 

[vanhees71]

I'm not sure what you want to calculate. The singlet state is
$$|\psi \rangle=\frac{1}{\sqrt{2}} (|1/2,-1/2 \rangle-|-1/2, 1/2 \rangle).$$
I suppose that's the state the spins are prepared in. You can calulate the expectation value of $\vec{s}_1 \cdot \vec{s}_2$ using this state. It's easier by using
$$(\vec{s}_1 + \vec{s}_2)^2=\vec{s}_1^2 + \vec{s}_2^2 +2 \vec{s}_1 \cdot \vec{s}_2.$$
Taking the average gives on the left-hand side simply 0, because the total spin is $S=0$ and on the right-hand side
$$0=2 \hbar^2 1/2(1/2+1)+2 \langle \vec{s}_1 \cdot \vec{s}_2 \rangle \; \Rightarrow \; \langle \vec{s}_1 \cdot \vec{s}_2 \rangle=-\frac{3}{4} \hbar^2.$$
Then you can calculate the probatities for the various possible outcomes when you measure the spin components of each particle in some arbitrary given direction by calculating the corresponding scalar products and take the modulus squared.

 

[allisonpw]

I agree with your final answer but am a bit puzzled over the numbers at the start of the last line. Elaborating on my original question- I imagine that in an entanglement experiment with the singlet state, the measurement of spin #1 is made at some unknown angle $\phi$ wrt to the orientation of your first equation. In this case the spin operator $\sigma_1$ along that axis is (I hope!)
$$ \sigma_1 = cos(\phi) \sigma_{z} + sin(\phi)\sigma_{x}$$
$$ =\begin{pmatrix}
cos(\phi)&sin(\phi)\\
sin(\phi)&-cos(\phi)
\end{pmatrix}$$
Measurement #2 is made at angle ($\phi+\theta$) , $\theta$ controlled by the experimenter, so $$ \sigma_2 =
\begin{pmatrix}
cos(\phi+\theta)&sin(\phi+\theta)\\
sin(\phi+\theta)&-cos(\phi+theta)
\end{pmatrix}$$

Then doing a page of algebra to evaluate $\langle\sigma_1\sigma_2\rangle$ I get (with $|\psi\rangle= (|ud\rangle-|du\rangle )/\sqrt2$)$$\langle\psi| \sigma_1\sigma_2|\psi\rangle = -cos(\theta)$$
Independent of $\phi$, this is what I would have expected from measurement comparisons with the Bell theorem that I've read about .

I then evaluated $\langle\sigma_1\sigma_2\rangle$ for the three cases that Susskind gives as the fully entangled triplet states, $$ (|ud\rangle + |du\rangle)/\sqrt2, (|uu\rangle \pm|dd\rangle)/\sqrt2$$ and got $-cos(\theta+2\phi), cos(\theta), +cos(\theta+2\phi)$ unexpectedly. The answers to this question must be well known! I'm suspicious that I've done something wrong.

 

[vanhees71]

The point is that the spin singlet has no direction. It's spin 0! That means the expectation value can only depend on the relative angle between the states. So your result makes at least qualitative sense. I've not checked it in detail.

 

[RUTA]

What you're computing is called the correlation function, it's the product of the two outcomes in each trial averaged over many trials. The easiest way to compute that, since the Bell basis states are all rotationally invariant, is to assume the state is given in the eigenbasis of $\sigma_1$. Then, $\sigma_2$ is your first matrix with $\theta$ where $\theta$ is the angle between the SG magnets represented by $\sigma_1$ and $\sigma_2$.

 

[vanhees71]

The $S=1$ states are not rotationally invariant but transform under the corresponding representation (the adjoint representation) of SU(2), which is the fundamental representation of SO(3).

 

[wle]

A nice and systematic way to do this kind of calculation is by writing the correlation in terms of the density operator, $$\langle \psi \rvert A \otimes B \lvert \psi \rangle = \text{Tr} \bigl[ (A \otimes B) \lvert \psi \rangle \langle \psi \rvert \bigr] \,,$$ and then expressing the density operator in terms of the Pauli operators. For example, for the singlet state $\lvert \psi_{-} \rangle = (\lvert 01 \rangle - \lvert 01 \rangle)/\sqrt{2}$, the corresponding density operator can be written as $$\lvert \psi_{-} \rangle \langle \psi_{-} \rvert = \frac{1}{4} \Bigl[ \mathbb{I} \otimes \mathbb{I} - \sigma_{x} \otimes \sigma_{x} - \sigma_{y} \otimes \sigma_{y} - \sigma_{z} \otimes \sigma_{z} \Bigr] \,,$$ where $\mathbb{I}$, $\sigma_{x}$, $\sigma_{y}$, and $\sigma_{z}$ are the identity and Pauli operators. You can easily derive expressions like this by first expanding the density operator as (in this case) $$\begin{align*}
\lvert \psi_{-} \rangle \langle \psi_{-} \rvert = \frac{1}{2} \Bigl( &\lvert 0 \rangle \langle 0 \rvert \otimes \lvert 1 \rangle \langle 1 \rvert + \lvert 1 \rangle \langle 1 \rvert \otimes \lvert 0 \rangle \langle 0 \rvert \\
&- \lvert 0 \rangle \langle 1 \rvert \otimes \lvert 1 \rangle \langle 0 \rvert - \lvert 1 \rangle \langle 0 \rvert \otimes \lvert 0 \rangle \langle 1 \rvert \Bigr)
\end{align*}$$ and then substituting in the inverses of the expressions $$\begin{eqnarray*}
\mathbb{I} &=& \lvert 0 \rangle \langle 0 \rvert + \lvert 1 \rangle \langle 1 \rvert \,, & \qquad
\sigma_{x} &=& \lvert 0 \rangle \langle 1 \rvert + \lvert 1 \rangle \langle 0 \rvert \,, \\
\sigma_{z} &=& \lvert 0 \rangle \langle 0 \rvert - \lvert 1 \rangle \langle 1 \rvert \,, & \qquad
\sigma_{y} &=& -i \lvert 0 \rangle \langle 1 \rvert + i \lvert 1 \rangle \langle 0 \rvert
\end{eqnarray*}$$ for the identity and Pauli operators. (For example, $\lvert 0 \rangle \langle 0 \rvert = \tfrac{1}{2} (\mathbb{I} + \sigma_{z})$.)

With this it's pretty easy to work out the correlation for arbitrary observables if you remember the properties (mainly, $\sigma_{i} \sigma_{j} = \delta_{ij} \mathbb{I} + i \varepsilon_{ijk} \sigma_{k}$, which implies $\text{Tr}[\sigma_{i} \sigma_{j}] = 2 \delta_{jk}$) of the Pauli operators. With $A = \boldsymbol{a} \cdot \boldsymbol{\sigma}$ and $B = \boldsymbol{b} \cdot \boldsymbol{\sigma}$ you get $$\begin{eqnarray*}
\langle \psi_{-} \rvert A \otimes B \lvert \psi_{-} \rangle &=&\text{Tr} \bigl[ \bigl((\boldsymbol{a} \cdot \boldsymbol{\sigma}) \otimes (\boldsymbol{b} \cdot \boldsymbol{\sigma}) \bigr) \lvert \psi_{-} \rangle \langle \psi_{-} \rvert \bigr] \\
&=& \frac{1}{4} a_{i} b_{j} \text{Tr} \Bigl[ (\sigma_{i} \otimes \sigma_{j}) \bigl( \mathbb{I} \otimes \mathbb{I} - \sigma_{x} \otimes \sigma_{x} - \sigma_{y} \otimes \sigma_{y} - \sigma_{z} \otimes \sigma_{z} \bigr) \Bigr] \\
&=& - \frac{1}{4} a_{i} b_{j} \text{Tr}[\sigma_{i} \sigma_{k}] \text{Tr}[\sigma_{j} \sigma_{k}] \\
&=& - a_{i} b_{j} \, \delta_{ik} \delta_{jk} \\
&=& - \boldsymbol{a} \cdot \boldsymbol{b} \,.
\end{eqnarray*}$$

The same method works just as well for other maximally-entangled states. For example, for $\lvert \phi_{+} \rangle = (\lvert 00 \rangle + \lvert 11 \rangle)/\sqrt{2}$, the corresponding density operator is $$\lvert \phi_{+} \rangle \langle \phi_{+} \rvert = \frac{1}{4} \Bigl[ \mathbb{I} \otimes \mathbb{I} +\sigma_{x} \otimes \sigma_{x} - \sigma_{y} \otimes \sigma_{y} + \sigma_{z} \otimes \sigma_{z} \Bigr]$$ and for the correlations you get $$\langle \phi_{+} \rvert (\boldsymbol{a} \cdot \boldsymbol{\sigma}) \otimes (\boldsymbol{b} \cdot \boldsymbol{\sigma}) \lvert \phi_{+} \rangle = \boldsymbol{a} \cdot \boldsymbol{b}^{*}$$ where $\boldsymbol{b}^{*} = (b_{x}, -b_{y}, b_{z})$ is the vector $\boldsymbol{b}$ except with its $y$ component flipped.

 

[allisonpw]

What you're computing is called the correlation function, it's the product of the two outcomes in each trial averaged over many trials. The easiest way to compute that, since the Bell basis states are all rotationally invariant, is to assume the state is given in the eigenbasis of $\sigma_1$. Then, $\sigma_2$ is your first matrix with $\theta$ where $\theta$ is the angle between the SG magnets represented by $\sigma_1$ and $\sigma_2$.

 

[allisonpw]

I'm closely following the postings to my question, and thanks to all who have replied! I hadn't previously considered whether any of the entangled states would be rotationally invariant, so I applied a rotation to all four and found that the singlet, S= $(|ud \rangle - |du \rangle )/\sqrt2$ and one of the symmetric states, T2 = $(|uu \rangle + |dd \rangle)/\sqrt2$ were indeed invariant, but the other two symmetric states were not. Perhaps this is in agreement with vanhees71's comment. Since $\langle\sigma\rangle$ = 0 in all four cases, $\langle\sigma_1\sigma_2\rangle$ is the same as the correlation. I'll be on vacation next week.

 

[RUTA]

By "rotational invariance" I mean no matter what Alice and Bob are measuring, as long as they're measuring the same thing the Bell basis states are the same. When you're talking about $|\phi_+\rangle$ and $|\psi_-\rangle$ as in Post #7, then it's a physical rotation of the SG magnets in the plane orthogonal to the beam. The angle of the physical rotation of the SG magnets is $2\theta$ when the angle of rotated Hilbert space basis vectors preserving the functional form of $|\phi_+\rangle$ and $|\psi_-\rangle$ is $\theta$. This is a rotation of the SG magnets by $2\theta$ in the x-z plane, i.e., $\sigma_2 = \cos(2\theta)\sigma_z + \sin(2\theta)\sigma_x$.

 

[RUTA]

The transformation in Hilbert space that leaves $|\phi_-\rangle$ invariant is $|u\rangle \rightarrow \cos(\theta)|u\rangle + i\sin(\theta)|d\rangle$ and $|d\rangle \rightarrow i\sin(\theta)|u\rangle + \cos(\theta)|d\rangle$. Construct your operator from these states, i.e., $\sigma_2 = |u\rangle\langle u| - |d\rangle\langle d|$ as you did for $|\phi_+\rangle$ and $|\psi_-\rangle$, and you get $\sigma_2 = \cos(2\theta)\sigma_z + \sin(2\theta)\sigma_y$. So, you see this just means you're now rotating your SG magnets in the y-z plane instead of the x-z plane.

 

[wle]

More generally, the density operator associated to the $\lvert \psi_{-} \rangle$ state is invariant under any unitary transformation applied to both parts. The ket $\lvert \psi_{-} \rangle$ itself is invariant under any unitary in $\text{SU}(2)$ applied to both sides and invariant up to a phase for any unitary in $\text{U}(2)$. More precisely, $$U \otimes U \lvert \psi_{-} \rangle = \det(U) \lvert \psi_{-} \rangle$$ for any unitary $U$. (Actually, that relation holds for any operator $A$ applied on both sides, not just unitaries.) The other Bell states also have symmetries with respect to any unitary, but generally not with the same unitary applied to both sides. For example, for $\lvert \phi_{+} \rangle$ you have $$U \otimes U^{*} \lvert \phi_{+} \rangle = \lvert \phi_{+} \rangle$$ where $U$ is any unitary and $U^{*}$ is its complex conjugate with respect to the basis $\{\lvert 0 \rangle, \lvert 1 \rangle\}$.

In terms of rotations in the Bloch sphere representation (which you can think of as applying either to the observables or to the state itself), the symmetries are whatever leave the products such as appear in post #7 invariant. So $\lvert \psi_{-} \rangle$ is invariant if you apply any rotation as long as you do the same one on both sides, which leaves the scalar product $\boldsymbol{a} \cdot \boldsymbol{b}$ invariant. $\lvert \phi^{+} \rangle$ or correlations generated using it are invariant if you apply any rotation on one side and its mirror image through the $x-z$ plane on the other side (i.e., whatever you do in the $y$ direction on one side you do in the $-y$ direction on the other side), which leaves the product $\boldsymbol{a} \cdot \boldsymbol{b}^{*} = a_{x} b_{x} - a_{y} b_{y} + a_{z} b_{z}$ invariant. For the states $\lvert \phi_{-} \rangle = (\lvert 00 \rangle - \lvert 11 \rangle)/\sqrt{2}$ and $\lvert \psi_{+} \rangle = (\lvert 01 \rangle + \lvert 10 \rangle) / \sqrt{2}$ you have $$\begin{eqnarray*}
\langle \phi_{-} \rvert A \otimes B \lvert \phi_{-} \rangle &=& - a_{x} b_{x} + a_{y} b_{y} + a_{z} b_{z} \,, \\
\langle \psi_{+} \rvert A \otimes B \lvert \psi_{+} \rangle &=& a_{x} b_{x} + a_{y} b_{y} - a_{z} b_{z}
\end{eqnarray*}$$ which have similar symmetries, except with the mirroring through the $y-z$ and $x-y$ planes in those cases.

 

[RUTA]

Ok, so far I have shown that $|\psi_-\rangle$ (S = 0 singlet state) is rotationally invariant in the x-z plane (indeed, as @wle showed, its correlation function is a dot product in any plane). That means we can choose $\sigma_1 = \sigma_z$, i.e., the eigenbasis being used, and $\sigma_2 = \cos(2\theta)\sigma_z + \sin(2\theta)\sigma_x$. This gives a correlation function of $-\cos(2\theta)$ for $|\psi_-\rangle$ (from @wle $a_x = a_y = 0; a_z = 1$ and $b_z = \cos(2\theta)$). All this obtains because we have conservation of angular momentum with the total equal to zero. Now for the S = 1 triplet states.

I showed that $|\phi_+\rangle$ is also invariant under rotations in the x-z plane, so we can again choose $\sigma_1 = \sigma_z$ and $\sigma_2 = \cos(2\theta)\sigma_z + \sin(2\theta)\sigma_x$. This gives a correlation function of $\cos(2\theta)$ for $|\phi_+\rangle$ (from @wle $a_x = a_y = 0; a_z = 1$ and $b_z = \cos(2\theta)$). Restricting ourselves to measurements in the x-z plane means the correlation function is given by $\hat{a}\cdot\hat{b}$ in @wle's analysis, which is how his result shows rotational invariance in the x-z plane for $|\phi_+\rangle$.

Next, I showed that $|\phi_-\rangle$ is invariant under rotations in the y-z plane, so we can again choose $\sigma_1 = \sigma_z$ and $\sigma_2 = \cos(2\theta)\sigma_z + \sin(2\theta)\sigma_y$. This gives a correlation function of $\cos(2\theta)$ for $|\phi_-\rangle$ (from @wle $a_x = a_y = 0; a_z = 1$ and $b_z = \cos(2\theta)$). Restricting ourselves to measurements in the y-z plane means the correlation function is given by $\hat{a}\cdot\hat{b}$ in @wle's analysis, which is how his result shows rotational invariance in the y-z plane for $|\phi_-\rangle$.

To finish the analysis, we need to find the plane in which $|\psi_+\rangle$ is invariant under rotations. We look to @wle's analysis and realize that we have a correlation function of $\hat{a}\cdot\hat{b}$ for $|\psi_+\rangle$ when we're in the x-y plane, so first put $|\psi_+\rangle$ in the eigenbasis of $\sigma_x$ via rotation by $45^o$ and then look to find the transformation that leaves it invariant (in other words, get us out of the z basis, since we suspect that our plane of rotational invariance will be the x-y plane). We find $|\psi_+\rangle \rightarrow |\phi_-\rangle$ in the x basis and we already solved that problem. So, we now know that our plane of rotational invariance is the x-y plane and we can choose $\sigma_1 = \sigma_x$, i.e., our eigenbasis, and we have $\sigma_2 = \cos(2\theta)\sigma_x + \sin(2\theta)\sigma_y$. This then gives a correlation function of $\cos(2\theta)$ for $|\psi_-\rangle$ (from @wle $a_y = a_z = 0; a_x = 1$ and $b_x = \cos(2\theta)$). Restricting ourselves to measurements in the x-y plane means the correlation function is given by $\hat{a}\cdot\hat{b}$ in @wle's analysis, which is how his result shows rotational invariance in the x-y plane for $|\psi_+\rangle$.

What does all this mean? Obviously, each triplet state represents a plane containing your S = 1 conserved angular momentum vector. If you want your conserved S = 1 angular momentum vector to reside in another plane, you simply create a superposition, i.e., expand in the triplet basis. Now, all this sometimes leads people to believe that we actually have a hidden conserved pair of aligned vectors of value $\frac{\hbar}{2}$. But, since every state tells you that Alice and Bob will obtain $\pm \frac{\hbar}{2}$ for every measurement, it would be an amazing coincidence or contrivance if those outcomes were based on a hidden variable. Instead, see my Insight for what I believe is a much more reasonable explanation for this type of conservation.

 

[RUTA]

I've written an Insight detailing my calculation of the correlation function for the Bell states. I'll give a link to it here once Greg gets it posted.

 

[allisonpw]

Thanks to all who responded to my Oct 10 and 11 question, which has now been well answered, particularly by RUTA and wle #12. I find that my original calculation was correct but did not include all the possibilities. The triplet states have more complex features than I had thought!